Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;

public class Main
{
  static int n;
  static double[] p;
  static Double[][] memo;
  public static void main(String[] args)
  {
	Scanner in = new Scanner(System.in);
    n = in.nextInt();
    p = new double[n];
    for (int i = 0; i < n; i++)
      p[i] = in.nextDouble();
    memo = new Double[n][n];
    System.out.println(dp(0, 0));
    in.close();
  }
  
  public static double dp(int at, int cnt)
  {
	if (at == n)
      return cnt > n/2 ? 1.0 : 0.0;
    if (memo[at][cnt] != null)
      return memo[at][cnt];
    double ans = 0;
    ans = p[at]*dp(at+1, cnt+1)+(1-p[at])*dp(at+1, cnt);
    return memo[at][cnt] = ans;
 }
}